Binomial Distribution

Consider exercise 4.45 on page 255 of your textbook.

Suppose that a student is taking a multiple-choice exam in which each question
has four choices. Assuming that she has no knowledge of the correct to any of
the questions, she has decided on a strategy in which she will place four balls
(marked A, B, C, and D) into a box. She randomly selects one ball for each
question and replaces the ball in the box. The marking on the ball will
determine her answer to the question.

Using this method (not recommended for MA 216!), the probability of getting
a correct answer would be 1/4 = .25 for each question. This is a binomial
situation with n=5 and p=.25.

Note that all of the properties of the binomial distribution are satisfied:

1.

The student is sampling with replacement.

2.

There are only two outcomes: A correct answer (Success) or an
incorrect answer (Failure). We are not concerned with which wrong
answer it might be among the three possibilities.

Assuming the lists L1, L2, and L3 are empty, put the sample
space {0, 1, 2, 3, 4, 5} in L1, using the seq function under
LIST OPS. Simply insert "A" for the expression and variable,
since we just want the sequence from 0 to 5. Choose binompdf from
the DISTR menu, giving (5, .25) for the (n, p) arguments. Store this
in L2. Now choose binomcdf and supply the same arguments.
Store this in L3. Now go to STAT Edit to look at the lists.

(a)(2)
can be found either by adding
P(X=4) + P(X=5) from
L2, which is
.01465 + .00098 = .01563, or by taking
from L3.

(a)(3)
P(X=0) = .2373, from L2.

(a)(4)
,
from L3.

Part (d) can be done easily with the calculator.

(d) Now we have n=50, p=.25, and we want
.
Here, let's just
get the cdf list, since adding up the pdf column from 30 to 50 would be almost
as tedious as doing the problem by hand!
Put the sample space {0, 1, ..., 50} in L4. Put the cdf list in
L5. Again, go to STAT Edit to look at the lists. Scroll down
so that the "30" entry is in view.

.
So the probability is zero!
Not much chance of passing the exam with this strategy. Actually, this
probability isn't exactly zero, but it is very close to zero. The
TI-83 ran out of significant digits to display. This will be illustrated
when we use Excel/PHStat to do this problem next.

Visit www.stat.wmich.edu/s216/phstat.html to learn how to use the
Excel/PHStat software.

Choose Probability Distributions from the PHStat menu, then pick
Binomial.

For part (a), enter 5 for the Sample Size. The Probability
of Success should be .25, then enter 0 and 5 for Outcomes From:
and To: , respectively. Enter an Output Title if you wish, and
put a checkmark in the box next to Cumulative Probabilities.

On the Binomial sheet, notice that the mean and standard deviation are
given. These match the results we obtained by hand. Column B contains
the possible outcomes, C lists the pdf or P(X=x) probabilities (L2
when we were using the TI-83), D lists the cdf or
probabilities (L3 from before), and columns E, F, and G contain the
probabilities from other ways to list them. Slightly overkill, but we can
make use of column G in this problem.

(a)(1)
P(X=5) = .000977, from column C.

(a)(2)
,
from column G.

(a)(3)
P(X=0) = .237305, from column C.

(a)(4)
,
from column D.

(d) Re-do the Binomial Probability Distribution dialog box with
n=50, p=.25, Outcomes from 0 to 50, and be sure the
Cumulative Probabilities box is checked.